A hemispherical cavity of radius R is created in a solid sphere of radius 2R as shown in the figure. The y -coordinate of the centre of mass of the remaining sphere is

Question

A hemispherical cavity of radius R is created in a solid sphere of radius 2R as shown in the figure. The y -coordinate of the centre of mass of the remaining sphere is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-jcrub2swezqdawqn.jpg" /> (A) yCM=-(R/15) (B) yCM=-(R/40) (C) yCM=-(R/30) (D) yCM=-(R/20)

Tardigrade Admin · Accepted Answer

x CM =((4/3) π(2 R )3 0-(2/3) π R 3 R /(4)3 π(2 R )3-(2)3 π R 3=-( R /15) y / CM )=((4/3) π(2 R )3 0-(2/3) π R 3 (3 R /8)/(4)3 π(2 R )3-(2)3 π R 3=-( R /40)

Q. A hemispherical cavity of radius $R$ is created in a solid sphere of radius $2 R$ as shown in the figure. The $y$ -coordinate of the centre of mass of the remaining sphere is

$y_{CM} = - \frac{R}{15}$

$y_{CM} = - \frac{R}{40}$

$y_{CM} = - \frac{R}{30}$

$y_{CM} = - \frac{R}{20}$

$x_{CM} = \frac{\frac{4}{3} π ( 2 R ) ^{3} 0 - \frac{2}{3} π R ^{3} R}{\frac{4}{3} π ( 2 R ) ^{3} - \frac{2}{3} π R ^{3}} = - \frac{R}{15}$
$y_{CM} = \frac{\frac{4}{3} π ( 2 R ) ^{3} 0 - \frac{2}{3} π R ^{3} \frac{3 R}{8}}{\frac{4}{3} π ( 2 R ) ^{3} - \frac{2}{3} π R ^{3}} = - \frac{R}{40}$

Q. A hemispherical cavity of radius R is created in a solid sphere of radius 2R as shown in the figure. The y -coordinate of the centre of mass of the remaining sphere is

yCM​=−15R​

yCM​=−40R​

yCM​=−30R​

yCM​=−20R​